Sharp Coefficient Problems of Functions with Bounded Turnings Subordinated by Sigmoid Function

Abstract

This study deals with analytic functions with bounded turnings, defined in the disk $O_d=\left\{z:\left|z\right|<1\right\}$. These functions are subordinated by sigmoid function $\frac{2}{1+e^{-z}}$ and their class is denoted by ${\mathfrak{BT}}_{\mathfrak{Sg}}$. Sharp coefficient inequalities, including the third Hankel determinant for class ${\mathfrak{BT}}_{\mathfrak{Sg}}$, were investigated here. The same was also included for the logarithmic coefficients related to functions of the class ${\mathfrak{BT}}_{\mathfrak{Sg}}$.

1. Introduction and Definitions

Let the class of analytic functions in disk $O_d=\left\{z\in \mathbb{C}:\left|z\right|<1\right\}$ be denoted by the notation $H\left(\mathcal{D}\right)$ and suppose that $\mathfrak{A}$ is the sub-family of $H\left(\mathcal{D}\right)$ defined as follows.

$$\mathfrak{A}:=\left\{f\in H\left(\mathcal{D}\right):f\left(z\right)=\sum _{k=1}^{\infty }{a}_k{z}^k,\mathrm{with}{a}_1=1\right\}.$$

(1)

Moreover, all univalent functions from class $\mathfrak{A}$ are composed in a class, named as $\mathcal{S}$. For two functions $g_1,g_2\in H\left(\mathcal{D}\right)$, the function $g_1$ is said to be subordinated by $g_2$, mathematically denoted as $g_1\prec g_2$, if a regular function $\upsilon$ defined in $O_d$ exists with the property that $\upsilon \left(0\right)=0$ and $\left|\upsilon \left(z\right)\right|<1$ such that $f\left(z\right)=g\left(\upsilon \left(z\right)\right)$. Moreover, if $g_2$ is univalent in $O_d$, then the relation $g_1\prec g_2$ with $g_1\left(0\right)=g_2\left(0\right)$ implies that $g_1\left(O_d\right)\subset g_2\left(O_d\right)$. For details, see [1,2,3] and the references therein.

Although the univalent function theory was initiated in 1851, the coefficient conjecture, proposed by Bieberbach [4] in 1916 and laterally proved by de-Branges [5] in 1985, turned the theory into one of the emerging areas of potential research. During the era between 1916 and 1985, several researchers attempted to prove or disprove this conjecture, which resulted in the formation of many subclasses of the class $\mathcal{S}$ that are based on the geometry of image domains. The most studied and fundamental subclasses of $\mathcal{S}$ are ${\mathcal{S}}^{\ast }$ and $\mathcal{K}$, which contain starlike and convex functions, respectively. Ma and Minda [6], in 1992, introduced the following general form of the class ${\mathcal{S}}^{\ast }$:

$${\mathcal{S}}^{\ast }\left(\varphi \right):=\left\{f\in \mathfrak{A}:\frac{z{f}^{\prime }\left(z\right)}{f\left(z\right)}\prec \varphi \left(z\right)\right\},$$

where $\varphi$ is an analytic function with $\Re \varphi \left(z\right)>0$, $z\in O_d$. In addition, the function f maps $O_d$ onto a star-shaped domain with respect to $\varphi \left(0\right)=1$ and is symmetric about the real axis. With the variation in function $\varphi$, the class ${\mathcal{S}}^{\ast }\left(\varphi \right)$ generates several sub-families of ${\mathcal{S}}^{\ast }$, which include some of the ones listed below.

(i).

If we choose $\varphi \left(z\right)=\frac{1+Lz}{1+Mz}$ with $-1\le M<L\le 1$, then we obtain ${\mathcal{S}}^{\ast }[L,M]\equiv {\mathcal{S}}^{\ast }\left(\frac{1+Lz}{1+Mz}\right)$, which is described as the class of Janowski starlike functions studied in [7]. Moreover, ${\mathcal{S}}^{\ast }\left(\xi \right):={\mathcal{S}}^{\ast }[1-2\xi ,-1]$ is the famous starlike functions’ class of order $\xi$ with $0\le \xi <1$.

(ii).

The family ${\mathcal{S}}^{\ast }\left(\varphi \right)$ with $\varphi \left(z\right)=\sqrt{1+z}$ was developed by Sokól and Stankiewicz in [8], which maps the symmetric disk $O_d$ onto the region bounded by $|w^2-1|<1$.

(iii).

By choosing the function $\varphi \left(z\right)=1+{sinh}^{-1}z$, we obtain ${\mathcal{S}}_{\rho }^{\ast }:={\mathcal{S}}^{\ast }\left(1+{sinh}^{-1}z\right)$, which was recently introduced by Kumar and Arora [9]. In 2021, Barukab et al. [10] found the sharp upper bound of the third Hankel determinant for functions of the following class:

$${\mathcal{R}}_s=\left\{f\in \mathfrak{A}:f^{\prime }\left(z\right)\prec 1+{sinh}^{-1}z,z\in O_d\right\}.$$

Later, in 2022, Shi et al. [11] determined the sharp second-order Hankel determinant for the above class, but with logarithmic coefficients.

(iv).

By choosing $\varphi \left(z\right)=1+\frac{4}{5}z+\frac{1}{5}{z}^4$, the class ${\mathcal{S}}^{\ast }\left(\varphi \right)$ reduces to class ${\mathcal{S}}_{3l}^{\ast }$, which was studied by Gandhi et al. [12]. In 2022, Arif et al. [13] determined the sharp third-order Hankel determinant for functions of class ${\mathcal{S}}_{3l}^{\ast }$. Later, Shi et al. [14] determined the sharp second Hankel determinant of the same class with logarithmic coefficients.

(v).

Raza and Bano [15] and Alotaibi et al. [16] contributed the families ${\mathcal{S}}_{cos}^{\ast }:={\mathcal{S}}^{\ast }\left(cos\left(z\right)\right)$ and ${\mathcal{S}}_{cosh}^{\ast }:={\mathcal{S}}^{\ast }\left(cosh\left(z\right)\right)$, respectively. These researchers investigated some geometric characteristics of the functions of these families.

(vi).

We obtain the family ${\mathcal{S}}_{sin}^{\ast }$ by choosing $\varphi \left(z\right)=1+sinz$, which was established in [17]. In this paper, the authors determined radii problems for the defined class ${\mathcal{S}}_{sin}^{\ast }$.

For functions $f\in \mathcal{S}$ of the series form as given in (1), the Hankel determinant ${\mathsf{Ħ}}_{q,n}\left(f\right)$$\left(\mathrm{with}q,n\in \mathbb{N}=\left\{1,2,\dots \right\}\mathrm{and}{a}_1=1\right)$ was given by Pommerenke [18,19] and is defined as follows:

$${\mathsf{Ħ}}_{q,n}\left(f\right):=\left|\begin{array}{cccc}{a}_n\hfill & a_{n+1}\hfill & \dots \hfill & a_{n+q-1}\hfill \\ a_{n+1}\hfill & a_{n+2}\hfill & \dots \hfill & a_{n+q}\hfill \\ ⋮\hfill & ⋮\hfill & \dots \hfill & ⋮\hfill \\ a_{n+q-1}\hfill & a_{n+q}\hfill & \dots \hfill & a_{n+2q-2}\hfill \end{array}\right|.$$

With certain variations in q and n, we have the following second and third-order Hankel determinants, respectively:

$$\begin{array}{ccc}\hfill {\mathsf{Ħ}}_{2,1}\left(f\right)& =& \left|\begin{array}{cc}1\hfill & a_2\hfill \\ a_2\hfill & a_3\hfill \end{array}\right|=a_3-a_2^2,\hfill \\ \hfill {\mathsf{Ħ}}_{2,2}\left(f\right)& =& \left|\begin{array}{cc}{a}_2\hfill & a_3\hfill \\ a_3\hfill & a_4\hfill \end{array}\right|=a_2{a}_4-a_3^2,\hfill \\ \hfill {\mathsf{Ħ}}_{3,1}\left(f\right)& =& \left|\begin{array}{ccc}1\hfill & a_2\hfill & a_3\hfill \\ a_2\hfill & a_3\hfill & a_4\hfill \\ a_3\hfill & a_4\hfill & a_5\hfill \end{array}\right|=\left(a_2{a}_4-a_3^2\right)a_3-\left(a_4-a_2{a}_3\right)a_4+\left(a_3-a_2^2\right)a_5.\hfill \end{array}$$

(2)

For the functions $f\in \mathcal{S}$, the best established sharp inequality is $\left|{\mathsf{Ħ}}_{2,n}\left(f\right)\right|\le \lambda \sqrt{n}$, where $\lambda$ is an absolute constant and this inequality is due to Hayman [20]. Furthermore, the following results for the class $\mathcal{S}$ can be accessed from [21].

$$\begin{array}{ccc}\hfill \left|{\mathsf{Ħ}}_{2,2}\left(f\right)\right|& \le & \lambda ,\mathrm{for}1\le \lambda \le \frac{11}{3},\hfill \\ \hfill \left|{\mathsf{Ħ}}_{3,1}\left(f\right)\right|& \le & \mu ,\mathrm{for}\frac{4}{9}\le \mu \le \frac{32+\sqrt{285}}{15}.\hfill \end{array}$$

The following sharp bound of $\left|{\mathsf{Ħ}}_{2,2}\left(f\right)\right|$ was given by Janteng et al. [22,23].

$$\left|{\mathsf{Ħ}}_{2,2}\left(f\right)\right|\le \left\{\begin{array}{ccc}\frac{1}{8},\hfill & \mathrm{for}\hfill & f\in \mathcal{K},\hfill \\ 1,\hfill & \mathrm{for}\hfill & f\in {\mathcal{S}}^{\ast },\hfill \\ \frac{4}{9},\hfill & \mathrm{for}\hfill & f\in \mathcal{R},\hfill \end{array}\right.$$

where $\mathcal{R}$ denotes the class of functions with bounded turnings which is defined as

$$\mathcal{R}=\left\{f\in \mathcal{S}:f^{\prime }\left(z\right)\prec \frac{1+z}{1-z},z\in O_d\right\}.$$

To date, a number of researchers have contributed to the work on Hankel determinants and have achieved remarkable milestones. Some of the recent developments can be accessed from [24,25,26,27,28,29,30,31,32,33,34] and the references therein. The logarithmic function associated to function $f\in \mathcal{S}$ is defined as

$$log\frac{f\left(z\right)}{z}=2\sum _{k=1}^{\infty }{\gamma }_k{z}^k.$$

(3)

If $f\in \mathcal{S}$ assumes the series of the form given in (1), then (3) gives the following relations:

$$\begin{array}{ccc}\hfill {\gamma }_1& =& \frac{1}{2}{a}_2\hfill \end{array}$$

(4)

$$\begin{array}{ccc}\hfill {\gamma }_2& =& \frac{1}{2}\left(a_3-\frac{1}{2}{a}_2^2\right)\hfill \end{array}$$

(5)

$$\begin{array}{ccc}\hfill {\gamma }_3& =& \frac{1}{2}\left(a_4-a_2{a}_3+\frac{1}{3}{a}_2^3\right)\hfill \end{array}$$

(6)

$$\begin{array}{ccc}\hfill {\gamma }_4& =& \frac{1}{2}\left(a_5-a_2{a}_4+a_2^2{a}_3-\frac{1}{2}{a}_3^2-\frac{1}{4}{a}_2^4\right).\hfill \end{array}$$

(7)

Recently, in [35,36], Kowalczyk and Lecko introduced the following qth-order Hankel determinant ${\mathsf{Ħ}}_{q,n}\left(F_f/2\right)$ containing the logarithmic coefficients of f.

$${\mathsf{Ħ}}_{q,n}\left(F_f/2\right)=\left|\begin{array}{cccc}{\gamma }_n\hfill & {\gamma }_{n+1}\hfill & \dots \hfill & {\gamma }_{n+q-1}\hfill \\ {\gamma }_{n+1}\hfill & {\gamma }_{n+2}\hfill & \dots \hfill & {\gamma }_{n+q}\hfill \\ ⋮\hfill & ⋮\hfill & \dots \hfill & ⋮\hfill \\ {\gamma }_{n+q-1}\hfill & {\gamma }_{n+q}\hfill & \dots \hfill & {\gamma }_{n+2q-2}\hfill \end{array}\right|.$$

(8)

From above, one can easily deduce that

$$\begin{array}{ccc}\hfill {\mathsf{Ħ}}_{2,1}\left(F_f/2\right)& =& {\gamma }_1{\gamma }_3-{\gamma }_2^2,\hfill \end{array}$$

(9)

$$\begin{array}{ccc}\hfill {\mathsf{Ħ}}_{2,2}\left(F_f/2\right)& =& {\gamma }_2{\gamma }_4-{\gamma }_3^2.\hfill \end{array}$$

(10)

We now define the class of functions with bounded turnings and associated with the sigmoid function $\varphi \left(z\right)=\frac{2}{1+e^{-z}}$ with $\varphi \left(0\right)=1$ and $\Re \varphi \left(z\right)>0$ as follows:

$${\mathfrak{BT}}_{\mathfrak{Sg}}:=\left\{f\in \mathcal{S}:f^{\prime }\left(z\right)\prec \frac{2}{1+e^{-z}},z\in O_d\right\}.$$

(11)

We intended to find the sharp bound of $\left|{\mathsf{Ħ}}_{3,1}\left(f\right)\right|$ and $\left|a_3{a}_5-a_4^2\right|$ for the class ${\mathfrak{BT}}_{\mathfrak{Sg}}$. In addition, we investigated the sharp bounds of $\left|{\mathsf{Ħ}}_{2,1}\left(F_f/2\right)\right|$ and $\left|{\mathsf{Ħ}}_{2,2}\left(F_f/2\right)\right|$ for the class ${\mathfrak{BT}}_{\mathfrak{Sg}}$.

2. A Set of Lemmas

Definition 1.

A function $p\in \mathcal{P}$ if and only if, it has the series expansion

$$p\left(z\right)=1+\sum _{n=1}^{\infty }{c}_n{z}^n,\left(z\in O_d\right)$$

(12)

along with the $\Re p\left(z\right)\ge 0$ for $z\in O_d$.

Lemma 1.

If the function $p\in \mathcal{P}$ has the series representation as given in (12), then

$$\left|c_{n+k}-\lambda c_n{c}_k\right|\le 2max\left\{1,\left|2\lambda -1\right|\right\}=\left\{\begin{array}{ccc}2& \mathrm{for}& 0\le \lambda \le 1;\\ 2\left|2\lambda -1\right|& & \mathrm{otherwise}.\end{array}\right.$$

(13)

and

$$\left|c_n\right|\le 2\mathrm{for}n\ge 1.$$

(14)

The inequalities (13) and (14) are sharp and their proofs can be accessed from [37,38], respectively.

Lemma 2

([39]). If the function $p\in \mathcal{P}$ has the series representation as given in (12) and if $B\in \left[0,1\right]$ with $B\left(2B-1\right)\le D\le B$, then, we have

$$\left|c_3-2B{c}_1{c}_2+D{c}_1^3\right|\le 2.$$

(15)

Lemma 3.

If the function $p\in \mathcal{P}$ has the series representation as given in (12), then, for $x,\delta ,\rho \in \overline{O_d}=\left\{z\in \mathbb{C}:\left|z\right|\le 1\right\}$, we have

$$\begin{array}{ccc}\hfill 2c_2& =& x\left(4-c_1^2\right)+c_1^2,\hfill \end{array}$$

(16)

$$\begin{array}{ccc}\hfill 4c_3& =& -x^2{c}_1\left(4-c_1^2\right)+2x{c}_1\left(4-c_1^2\right)+2\left(1-{\left|x\right|}^2\right)\left(4-c_1^2\right)\delta +c_1^3,\hfill \\ \hfill 8c_4& =& \left[4x+\left(x^2-3x+3\right)c_1^2\right]x(4-c_1^2)-4(1-{\left|x\right|}^2)(4-c_1^2)\hfill \end{array}$$

(17)

$$\begin{array}{ccc}& & -\rho (1-{\left|\delta \right|}^2)+(x-1)\delta c+{\delta }^2\overline{x}+c_1^4.\hfill \end{array}$$

(18)

Here, the readers can refer to the formula for $c_2$ given in [37]. The formula for $c_3$ is due to Libera and Zlotkiewicz [40], and the formula for $c_4$ is proved in [41].

Lemma 4

([42]). If the function $p\in \mathcal{P}$ has the series representation as given in (12) and if $\eta ,a,\alpha$ and β satisfy the inequalities $0<\alpha <1$, $0<a<1$, and

$$8\left({\left(\alpha \left(a+\alpha \right)-\beta \right)}^2+{\left(\alpha \beta -2\eta \right)}^2\right)\left(1-a\right)a+\alpha {\left(\beta -2a\alpha \right)}^2\left(1-\alpha \right)\le 4a{\alpha }^2{\left(1-\alpha \right)}^2\left(1-a\right),$$

then,

$$\left|\eta c_1^4+a{c}_2^2+2\alpha c_1{c}_3-\frac{3}{2}\beta c_1^2{c}_2-c_4\right|\le 2.$$

(19)

3. Third Hankel Determinant for the Class ${\mathfrak{BT}}_{\mathfrak{Sg}}$

Theorem 1.

If the function $f\in {\mathfrak{BT}}_{\mathfrak{Sg}}$ assumes the series representation as given in (1), then

$$\left|{\mathsf{Ħ}}_{3,1}\left(f\right)\right|\le \frac{1}{64}.$$

The inequality is sharp and sharpness can be achieved from

$$f\left(z\right)={\int }_0^z\left(\frac{2}{1+e^{-\left(t^3\right)}}\right)dt=z+\frac{z^4}{8}-\frac{z^{10}}{240}+\cdots .$$

Proof. 

Let $f\in {\mathfrak{BT}}_{\mathfrak{Sg}}$. Then,

$$f^{\prime }\left(z\right)=\frac{2}{1+e^{-w\left(z\right)}}.$$

(20)

If $p\in \mathcal{P}$, then

$$p\left(z\right)=\frac{1+\left(w\left(z\right)\right)}{1-\left(w\left(z\right)\right)}=1+c_{1}z+c_2{z}^2+c_3{z}^3+\cdots .$$

This implies that

$$\begin{array}{ccc}\hfill w\left(z\right)& =& \left(\frac{1}{2}{c}_1\right)z+\left(-\frac{1}{4}{c}_1^2+\frac{1}{2}{c}_2\right)z^2+\left(-\frac{1}{2}{c}_1{c}_2+\frac{1}{8}{c}_1^3+\frac{1}{2}{c}_3\right)z^3\hfill \\ & & +\left(\frac{3}{8}{c}_1^2{c}_2+\frac{1}{2}{c}_4-\frac{1}{4}{c}_2^2-\frac{1}{2}{c}_1{c}_3-\frac{1}{16}{c}_1^4\right)z^4+\cdots .\hfill \end{array}$$

(21)

Using (21) in (20), we obtain

$$\begin{array}{ccc}\hfill f^{\prime }\left(z\right)& =& 1+\frac{1}{4}{c}_{1}z+\left(-\frac{1}{8}{c}_1^2+\frac{1}{4}{c}_2\right)z^2+\left(\frac{1}{4}{c}_3-\frac{1}{4}{c}_2{c}_1+\frac{11}{192}{c}_1^3\right)z^3\hfill \\ & & +\left(\frac{11}{64}{c}_1^2{c}_2-\frac{1}{8}{c}_2^2-\frac{1}{4}{c}_1{c}_3-\frac{3}{128}{c}_1^4+\frac{1}{4}{c}_4\right)z^4+\cdots .\hfill \end{array}$$

(22)

From the series defined in (1), it follows that

$$f^{\prime }\left(z\right)=1+2a_{2}z+3a_3{z}^2+4a_4{z}^3+\cdots .$$

(23)

By comparing (22) and (23), one may have

$$\begin{array}{ccc}\hfill a_2& =& \frac{c_1}{8},\hfill \end{array}$$

(24)

$$\begin{array}{ccc}\hfill a_3& =& -\frac{c_1^2}{24}+\frac{c_2}{12}\hfill \end{array}$$

(25)

$$\begin{array}{ccc}\hfill a_4& =& \frac{11c_1^3}{768}+\frac{c_3}{16}-\frac{c_1{c}_2}{16},\hfill \end{array}$$

(26)

$$\begin{array}{ccc}\hfill a_5& =& -\frac{3c_1^4}{640}+\frac{11c_1^2{c}_2}{320}+\frac{c_4}{20}-\frac{c_1{c}_3}{20}-\frac{c_2^2}{40}.\hfill \end{array}$$

(27)

Substituting (24)–(27); in (2) and setting $c_1=c$, we obtain

$$\begin{array}{ccc}\hfill {\mathsf{Ħ}}_{3,1}\left(f\right)& =& \frac{1}{L}\left(-25344c^2{c}_4-384c^2{c}_2^2-23552c_2^3+3744c^3{c}_3-119c^6\right.\hfill \\ & & \left.-480c^4{c}_2+43776c{c}_2{c}_3-34560c_3^2+36864c_2{c}_4\right),\hfill \end{array}$$

(28)

where L = 8,847,360. Now, assuming $c_1=c$ and $m=\left(4-c_1^2\right)$ in (16)–(18), we obtain the following:

$$\begin{array}{ccc}\hfill c_2& =& \frac{1}{2}\left(mx+c^2\right),\hfill \\ \hfill c_3& =& \frac{1}{4}\left\{-m{x}^{2}c+2mxc+2m\left(1-{\left|x\right|}^2\right)\delta +c^3\right\},\hfill \\ \hfill c_4& =& \frac{1}{8}\left[\left(4x+\left(x^2-3x+3\right)c^2\right)mx-4m\left(1-{\left|x\right|}^2\right)\right.\hfill \\ & & \left.-\rho (1-{\left|\delta \right|}^2)+(x-1)\delta c+{\delta }^2\overline{x}+c^4\right].\hfill \end{array}$$

By inserting the above expressions into (28), one may have

$$\begin{array}{ccc}\hfill {\mathsf{Ħ}}_{3,1}\left(f\right)& =& \frac{1}{L}\left\{-3456\left(1-{\left|\delta \right|}^2\right)\left(1-{\left|x\right|}^2\right)c^{2}m\rho -15c^6+720c^{3}m\left(1-{\left|x\right|}^2\right)\delta \right.\hfill \\ & & +2880\left(1-{\left|x\right|}^2\right)cx{m}^2\delta +9216\left(1-{\left|\delta \right|}^2\right)\left(1-{\left|x\right|}^2\right)m^{2}x\rho -2944m^3{x}^3\hfill \\ & & +3456\left(1-{\left|x\right|}^2\right)c^{3}mx\delta +504c^4{x}^{2}m+3456\left(1-{\left|x\right|}^2\right)c^{2}m{\delta }^2\overline{x}\hfill \\ & & +96c^{4}mx-9216\left(1-{\left|x\right|}^2\right)m^{2}x\overline{x}{\delta }^2-8640m^2{\left(1-{\left|x\right|}^2\right)}^2{\delta }^2\hfill \\ & & +9216m^2{x}^3+288c^2{m}^2{x}^2-3456c^{2}m{x}^2+144c^2{m}^2{x}^4\hfill \\ & & \left.-3744c^2{m}^2{x}^3-864c^{4}m{x}^3-576c{m}^2{x}^2\left(1-{\left|x\right|}^2\right)\delta \right\}.\hfill \end{array}$$

Since $m=\left(4-c^2\right)$, it follows that

$${\mathsf{Ħ}}_{3,1}\left(f\right)=\frac{1}{L}\left({\mathfrak{m}}_0\left(c,x\right)+{\mathfrak{m}}_1\left(c,x\right)\delta +{\mathfrak{m}}_2\left(c,x\right){\delta }^2+\Pi \left(c,x,\delta \right)\rho \right),$$

where $\rho$, x, $\delta \in \overline{O_d}$, and

$$\begin{array}{ccc}\hfill {\mathfrak{m}}_0\left(c,x\right)& =& -15c^6-8x\left[2x\left(50x{c}^2-9x^2{c}^2+160x-18c^2\right)\left(4-c^2\right)\right.\hfill \\ & & \left.+432x{c}^2-12c^4+108c^4{x}^2-63c^{4}x\right]\left(4-c^2\right),\hfill \\ \hfill {\mathfrak{m}}_1\left(c,x\right)& =& -144c\left[\left(4-c^2\right)\left(-20x+4x^2\right)-5c^2-24x{c}^2\right]\left(1-{\left|x\right|}^2\right)\left(4-c^2\right),\hfill \\ \hfill {\mathfrak{m}}_2\left(c,x\right)& =& -576\left[\left(4-c^2\right)\left(15+{\left|x\right|}^2\right)-6c^2\overline{x}\right]\left(1-{\left|x\right|}^2\right)\left(4-c^2\right),\hfill \\ \hfill \Pi \left(c,x,\delta \right)& =& 1152\left(4-c^2\right)\left(1-{\left|x\right|}^2\right)\left(1-{\left|\delta \right|}^2\right)\left[8x\left(4-c^2\right)-3c^2\right].\hfill \end{array}$$

By replacing $\left|\delta \right|$ with y and $\left|x\right|$ with x, if we apply the statement $\left|\rho \right|\le 1$, it follows that

$$\begin{array}{ccc}\hfill \left|{\mathsf{Ħ}}_{3,1}\left(f\right)\right|& \le & \frac{1}{L}\left(\left|{\mathfrak{m}}_0\left(c,x\right)\right|+\left|{\mathfrak{m}}_1\left(c,x\right)\right|y+\left|{\mathfrak{m}}_2\left(c,x\right)\right|y^2+\left|\Pi \left(c,x,\delta \right)\right|\right),\hfill \\ & \le & \frac{1}{L}\left(\mathfrak{f}\left(c,x,y\right)\right).\hfill \end{array}$$

(29)

where

$$\mathfrak{f}\left(c,x,y\right)={\mathfrak{n}}_0\left(c,x\right)+{\mathfrak{n}}_1\left(c,x\right)y+{\mathfrak{n}}_2\left(c,x\right)y^2+{\mathfrak{n}}_3\left(c,x\right)\left(1-y^2\right).$$

(30)

with

$$\begin{array}{ccc}\hfill {\mathfrak{n}}_0\left(c,x\right)& =& 15c^6+8x\left[2x\left(50x{c}^2+9x^2{c}^2+160x+18c^2\right)\left(4-c^2\right)\right.\hfill \\ & & \left.+432x{c}^2+12c^4+108c^4{x}^2+63c^{4}x\right]\left(4-c^2\right),\hfill \\ \hfill {\mathfrak{n}}_1\left(c,x\right)& =& 144c\left[\left(4-c^2\right)\left(20x+4x^2\right)+5c^2+24x{c}^2\right]\left(1-x^2\right)\left(4-c^2\right),\hfill \\ \hfill {\mathfrak{n}}_2\left(c,x\right)& =& 576\left[\left(4-c^2\right)\left(15+x^2\right)+6c^{2}x\right]\left(1-x^2\right)\left(4-c^2\right),\hfill \\ \hfill {\mathfrak{n}}_3\left(c,x\right)& =& 1152\left[3c^2+\left(4-c^2\right)8x\right]\left(1-x^2\right)\left(4-c^2\right).\hfill \end{array}$$

Now, we have to maximize $\mathfrak{f}\left(c,x,y\right)$ in the closed cuboid $\Delta :\left[0,2\right]\times \left[0,1\right]\times \left[0,1\right]$. For this purpose, we need to find $max\mathfrak{f}\left(c,x,y\right)$ in the interior of $\Delta$, in the interior of all of its six faces, and on the twelve edges of cuboid $\Delta$.

(I). Initially, we will look for the maxima of $\mathfrak{f}\left(c,x,y\right)$ in the interior of $\Delta$.

Let $\left(c,x,y\right)\in \left(0,2\right)\times \left(0,1\right)\times \left(0,1\right)$. Then, on differentiating (30) partially with respect to the parameter y, it yields

$$\begin{array}{ccc}\hfill \frac{\partial \mathfrak{f}}{\partial y}& =& \left(1-x^2\right)\left(4-c^2\right)144\left[\left(8\left(x-15\right)\left(4-c^2\right)+48c^2\right)y\left(x-1\right)\right.\hfill \\ & & \left.+c\left(4x\left(4-c^2\right)\left(15+x\right)+24c^2\left(x+5\right)\right)\right].\hfill \end{array}$$

Taking $\frac{\partial \mathfrak{f}}{\partial y}=0$, we obtain

$$y=\frac{c\left(4x\left(4-c^2\right)\left(15+x\right)+24c^2\left(x+5\right)\right)}{\left(8\left(15-x\right)\left(4-c^2\right)-48c^2\right)\left(x-1\right)}:=y_1.$$

For $y_1$ to belong to $(0,1)$, it is possible only if

$$24c^3\left(x+5\right)+4cx\left(x+15\right)\left(4-c^2\right)+8\left(1-x\right)\left(15-x\right)\left(4-c^2\right)<48\left(1-x\right)c^2$$

(31)

and

$$c^2>\frac{4\left(120-8x\right)}{168-8x}.$$

(32)

Now, in order to find the solutions that meet both inequalities (31) and (32), we consider $v\left(x\right)=\frac{4\left(120-8x\right)}{168-8x}$. It follows that $v^{\prime }\left(x\right)<0$ in $\left(0,1\right)$; therefore, $v\left(x\right)$ is decreasing over $\left(0,1\right)$. Hence, $c^2>\frac{14}{5}$ and an easy calculation illustrates that (31) does not hold for $x\in \left(0,1\right)$. This implies that $\mathfrak{f}$ does not have critical point in $\left(0,2\right)\times \left(0,1\right)\times \left(0,1\right)$.

(II). Next, we will look for the maxima of $\mathfrak{f}\left(c,x,y\right)$ in the interior of all six faces of $\Delta$.

Choosing $c=0$, we achieve

$$\begin{array}{ccc}\hfill {\mathfrak{p}}_1\left(x,y\right)& =& \mathfrak{f}\left(0,x,y\right)=40960x^3+\left(9216\left(x-1\right)\left(x-15\right)y^2\right.\hfill \\ & & \left.+147456x\right)\left(1-x^2\right).\hfill \end{array}$$

Hence, we have found no maxima for $\mathfrak{f}\left(0,x,y\right)$ in $\left(0,1\right)\times \left(0,1\right)$. Taking $c=2$, we obtain

$$\mathfrak{f}\left(2,x,y\right)=960.$$

(33)

Setting $x=0$, we have

$$\begin{array}{ccc}\hfill {\mathfrak{p}}_2\left(c,y\right)& =& \mathfrak{f}\left(c,0,y\right)=720c^{3}y\left(4-c^2\right)+15c^6-3456c^4+12096c^4{y}^2\hfill \\ & & +13824c^2-82944c^2{y}^2+138240y^2.\hfill \end{array}$$

By solving $\frac{\partial {\mathfrak{p}}_2}{\partial y}=0$ and $\frac{\partial {\mathfrak{p}}_2}{\partial c}=0$, we obtain the critical points. Setting $\frac{\partial {\mathfrak{p}}_2}{\partial y}=0$, we achieve

$$y=\frac{5c^3}{24\left(7c^2-20\right)}:=y_0.$$

(34)

For the provided range of y, $y_0$ would belong to $\left(0,1\right)$ if $c>c_0$ for $c_0\approx 1.73573$. Further, $\frac{\partial {\mathfrak{p}}_2}{\partial c}=0$ gives

$$2160c^{2}y\left(4-c^2\right)-1440c^{4}y+90c^5-13824c^3+48384c^3{y}^2+27648c-165888c{y}^2=0.$$

By inserting (34) into the above equation, we obtain

$$1260c^9-682176c^7+5225472c^5-13271040c^3+11059200c=0.$$

Now, solving for $c\in \left(0,2\right)$, we achieve $c\approx 1.402523$. Hence, no optimal solution is achieved for $\mathfrak{f}\left(c,0,y\right)$ in $\left(0,2\right)\times \left(0,1\right)$. If we choose $x=1$, we find that

$${\mathfrak{p}}_3\left(c,y\right)=\mathfrak{f}\left(c,1,y\right)=-217c^6-4896c^4+13056c^2+40960.$$

(35)

For the critical point, $\frac{\partial {\mathfrak{p}}_3}{\partial c}=0$ gives $c_0\approx 1.110119$, at which, ${\mathfrak{p}}_3$ attains its maxima, which is

$${\mathfrak{p}}_3\left(c,y\right)\le 49207.94534.$$

Taking $y=0$, we obtain

$$\begin{array}{ccc}\hfill \mathfrak{f}\left(c,x,0\right)& =& {\mathfrak{p}}_4\left(c,x\right)=144c^6{x}^4+15c^6-64c^6{x}^3-1152c^4{x}^4-216c^6{x}^2\hfill \\ & & +2304c^2{x}^4-96c^{6}x-288c^4{x}^2+9600c^{4}x-9600c^4{x}^3\hfill \\ & & -3456c^4+4608c^2{x}^2+66048c^2{x}^3-73728c^{2}x\hfill \\ & & +13824c^2-106496x^3+147456x.\hfill \end{array}$$

A computation indicates that the solution for the system of equations

$$\frac{\partial {\mathfrak{p}}_4}{\partial c}=0\&\frac{\partial {\mathfrak{p}}_4}{\partial x}=0$$

in $\left(0,2\right)\times \left(0,1\right)$ does not exist. Considering $y=1$, we have

$$\begin{array}{ccc}\hfill {\mathfrak{p}}_5\left(c,x\right)& =& \mathfrak{f}\left(c,x,1\right)=-64c^6{x}^3+15c^6-576c^5{x}^4+144c^6{x}^4+576c^5{x}^3\hfill \\ & & -1728c^4{x}^4-216c^6{x}^2+6912c^2{x}^4+9216c^3{x}^3+82944c^2{x}^2\hfill \\ & & +1296c^5{x}^2-96c^{6}x+3072c^4{x}^3-11808c^4{x}^2-576c^{5}x+4608c^3{x}^4\hfill \\ & & -3072c^{4}x-21504c^2{x}^3-7488c^3{x}^2+8640c^4-9216c{x}^4-720c^5\hfill \\ & & -9216c^{3}x-69120c^2+46080cx-129024x^2+138240-9216x^4\hfill \\ & & -46080c{x}^3+2880c^3+13824c^{2}x+9216c{x}^2+40960x^3.\hfill \end{array}$$

From the computation, we conclude that the solution for the system of equations

$$\frac{\partial {\mathfrak{p}}_5}{\partial x}=0\&\frac{\partial {\mathfrak{p}}_5}{\partial c}=0$$

in $\left(0,2\right)\times \left(0,1\right)$ does not exist.

(III). Finally, we look forward to the maximum of $\mathfrak{f}\left(c,x,y\right)$ at the twelve edges of $\Delta$.

By substituting $y=0$ and $x=0$, it yields

$${\mathfrak{p}}_6\left(c\right)=\mathfrak{f}\left(c,0,0\right)=15c^6-3456c^4+13824c^2.$$

For the critical point, the equation $\frac{\partial {\mathfrak{p}}_6}{\partial c}=0$ gives $c_0=1.4236371$, at which, the maximum value is achieved for ${\mathfrak{p}}_6\left(c\right)$. That is,

$$\mathfrak{f}\left(c,0,0\right)\le 13946.406.$$

By selecting $y=1$ and $x=0$, we have

$${\mathfrak{p}}_7\left(c\right)=\mathfrak{f}\left(c,0,1\right)=-720c^5+15c^6+2880c^3+8640c^4+138240-69120c^2.$$

As $\frac{\partial {\mathfrak{p}}_7}{\partial c}<0$ for $\left[0,2\right]$, ${\mathfrak{p}}_7\left(c\right)$ is decreasing and achieves its maximum at $c=0$. Thus,

$$\mathfrak{f}\left(c,0,1\right)\le 138240.$$

By taking $x=0$ and $c=0$, we obtain

$${\mathfrak{p}}_8\left(y\right)=\mathfrak{f}\left(0,0,y\right)=138240y^2.$$

Clearly, $\frac{\partial {\mathfrak{p}}_8}{\partial y}>0$ for $\left[0,1\right]$, which indicates that ${\mathfrak{p}}_8\left(y\right)$ is increasing over $\left[0,1\right]$ and that its maximum value is attained at $y=1$. Thus, one may have

$$\mathfrak{f}\left(0,0,y\right)\le 138240.$$

We note that (35) is free of y. It follows that

$$\begin{array}{ccc}\hfill {\mathfrak{p}}_9\left(c\right)& =& \mathfrak{f}\left(c,1,0\right)=\mathfrak{f}\left(c,1,1\right).\hfill \\ \hfill {\mathfrak{p}}_9\left(c\right)& =& -217c^6-4896c^4+13056c^2+40960.\hfill \end{array}$$

For the critical point, the equation $\frac{\partial {\mathfrak{p}}_9}{\partial c}=0$ gives $c_0=1.110119$, at which, the maximum value of ${\mathfrak{p}}_9\left(c\right)$ is attained. Thus, one may conclude that

$${\mathfrak{p}}_9\left(c\right)\le 49207.94534.$$

For $x=1$ and $c=0$, we obtain

$${\mathfrak{p}}_{10}\left(y\right)=\mathfrak{f}\left(0,1,y\right)=40960.$$

By choosing $c=2$, we see that (33) is free of y, x, and c. It follows that

$$\mathfrak{f}\left(2,0,y\right)=\mathfrak{f}\left(2,1,y\right)=\mathfrak{f}\left(2,x,1\right)=\mathfrak{f}\left(2,x,0\right)=960.$$

By substituting $y=0$ and $c=0$, we have

$${\mathfrak{p}}_{11}\left(x\right)=\mathfrak{f}\left(0,x,0\right)=-106496x^3+147456x.$$

For the critical point, the equation $\frac{\partial {\mathfrak{p}}_{11}}{\partial x}=0$ gives $x_0=0.679366$, at which, the maximum value of ${\mathfrak{p}}_{11}\left(x\right)$ is attained. Hence, we have

$$\mathfrak{f}\left(0,x,0\right)\le \frac{98304}{13}\sqrt{6}\sqrt{13}.$$

By setting $y=1$ and $c=0$, we obtain

$${\mathfrak{p}}_{12}\left(x\right)=\mathfrak{f}\left(0,x,1\right)=-9216x^4+40960x^3-129024x^2+138240.$$

By simple computation, we see that ${\mathfrak{p}}_{12}\left(x\right)$ obtains its maximum value at 0, so we have

$$\mathfrak{f}\left(0,x,1\right)\le 138240.$$

Hence, from the above situations, we conclude that

$$\mathfrak{f}\left(c,x,y\right)\le 138240\mathrm{on}\Delta :\left[0,2\right]\times \left[0,1\right]\times \left[0,1\right].$$

By using Equation (29), it follows that

$${\mathsf{Ħ}}_{3,1}\left(f\right)\le \frac{1}{L}\left(\mathfrak{f}\left(c,x,y\right)\right)\le \frac{1}{64}.$$

Thus, the required result is accomplished. □

Theorem 2.

If the function $f\in {\mathfrak{BT}}_{\mathfrak{Sg}}$ assumes the series representation as given in (1), then

$$\left|a_3{a}_5-a_4^2\right|\le \frac{1}{64}.$$

The result is the best possible and equality is attained from the function

$$f\left(z\right)={\int }_0^z\left(\frac{2}{1+e^{-\left(t^3\right)}}\right)dt=z+\frac{z^4}{8}-\frac{z^{10}}{240}+\cdots .$$

Proof. 

Substituting (25)–(27) with $c_1=c$, we have

$$\begin{array}{ccc}\hfill a_3{a}_5-a_4^2& =& \frac{1}{2949120}\left(-29c^6-96c^4{c}_2+864c^3{c}_3-6144c^2{c}_4+10752c{c}_2{c}_3\right.\hfill \\ & & \left.-6144c_2^3+12288c_2{c}_4-11520c_3^2\right).\hfill \end{array}$$

(36)

Using the resulting form of (16)–(18) with $m=\left(4-c^2\right)$, we obtain

$$\begin{array}{ccc}\hfill 10752c{c}_2{c}_3& =& -1344c^2{m}^2{x}^3-1344c^{4}m{x}^2+2688cx{m}^2\left(1-{\left|x\right|}^2\right)\delta +2688c^2{m}^2{x}^2\hfill \\ & & +2688c^{3}m\left(1-{\left|x\right|}^2\right)\delta +1344c^6+4032c^{4}mx,\hfill \\ \hfill 96c^4{c}_2& =& 48c^{4}mx+48c^6,\hfill \\ \hfill 6144c_2^3& =& 768\left(m^3{x}^3+3c^2{m}^2{x}^2+3c^{4}mx+c^6\right),\hfill \\ \hfill 6144c^2{c}_4& =& 768\left(4\left(1-{\left|\delta \right|}^2\right)\left(1-{\left|x\right|}^2\right)c^{2}m\rho +c^{4}m{x}^3-3c^{4}m{x}^2-4\left(1-{\left|x\right|}^2\right)c^{3}xm\delta \right.\hfill \\ & & \left.+3c^{2}a{x}^2+4\left(1-{\left|x\right|}^2\right)c^3\delta m+3c^{4}mx-4\left(1-{\left|x\right|}^2\right)c^2{\delta }^{2}m\overline{x}+c^6\right),\hfill \\ \hfill 12288c_2{c}_4& =& 768\left(4c^{2}m{x}^2+4m^2{x}^3+c^6+4c^{4}mx+4c^{3}m\left(1-{\left|x\right|}^2\right)\delta -3c^{4}m{x}^2\right.\hfill \\ & & -4\left(1-{\left|x\right|}^2\right)c^{3}m\delta x+4c^{2}m\left(1-{\left|x\right|}^2\right)\left(1-{\left|\delta \right|}^2\right)\rho +4c{m}^{2}x\left(1-{\left|x\right|}^2\right)\delta \hfill \\ & & -3c^2{m}^2{x}^3-4\left(1-{\left|x\right|}^2\right)c^2\overline{x}{\delta }^{2}m+c^{4}m{x}^3+4\left(1-{\left|\delta \right|}^2\right)\left(1-{\left|x\right|}^2\right)m^2\rho x\hfill \\ & & \left.+3c^2{m}^2{x}^2-4\left(1-{\left|x\right|}^2\right)m^2{x}^2\delta c+c^2{m}^2{x}^4-4\left(1-{\left|x\right|}^2\right)x\overline{x}{\delta }^2{m}^2\right),\hfill \\ \hfill 864c^3{c}_3& =& 432c^{4}mx+432\left(1-{\left|x\right|}^2\right)c^{3}m\delta +216c^6-216c^{4}m{x}^2,\hfill \\ \hfill 11520c_3^2& =& 720\left(c^2{m}^2{x}^4-4c{m}^2{x}^2\left(1-{\left|x\right|}^2\right)\delta -4c^2{m}^2{x}^3-2c^{4}m{x}^2+c^6+4c^{4}mx\right.\hfill \\ & & \left.+8c{m}^{2}x\left(1-{\left|x\right|}^2\right)\delta +4c^{3}m\left(1-{\left|x\right|}^2\right)\delta +4c^2{m}^2{x}^2+4m^2{\left(1-{\left|x\right|}^2\right)}^2{\delta }^2\right).\hfill \end{array}$$

By inserting the above expressions into (36), one may have

$$\begin{array}{ccc}\hfill a_3{a}_5-a_4^2& =& \frac{1}{2949120}\left\{-5c^6+240c^{3}m\left(1-{\left|x\right|}^2\right)\delta -120c^4{x}^{2}m-192c^2{m}^2{x}^2\right.\hfill \\ & & -2880m^2{\left(1-{\left|x\right|}^2\right)}^2{\delta }^2-768m^3{x}^3+3072x^3{m}^2+48c^2{m}^2{x}^4\hfill \\ & & -192c{x}^2{m}^2\left(1-{\left|x\right|}^2\right)\delta +3072m^{2}x\left(1-{\left|x\right|}^2\right)\left(1-{\left|\delta \right|}^2\right)\rho \hfill \\ & & \left.-3072x{m}^2\overline{x}\left(1-{\left|x\right|}^2\right){\delta }^2-768c^2{m}^2{x}^3\right\}.\hfill \end{array}$$

Since $m=4-c^2$, it follows that

$$a_3{a}_5-a_4^2=\frac{1}{2949120}\left({\mathfrak{b}}_0\left(c,x\right)+{\mathfrak{b}}_1\left(c,x\right)\delta +{\mathfrak{b}}_2\left(c,x\right){\delta }^2+{\mathfrak{b}}_3\left(c,x,\delta \right)\rho \right),$$

where $\delta ,\rho ,x\in \overline{O_d}$, and

$$\begin{array}{ccc}\hfill {\mathfrak{b}}_0\left(c,x\right)& =& -5c^6-24x^2\left[5c^4+2\left(4c^2-x^2{c}^2\right)\left(4-c^2\right)\right]\left(4-c^2\right),\hfill \\ \hfill {\mathfrak{b}}_1\left(c,x\right)& =& -48c\left[-5c^2+4x^2\left(4-c^2\right)\right]\left(1-{\left|x\right|}^2\right)\left(4-c^2\right),\hfill \\ \hfill {\mathfrak{b}}_2\left(c,x\right)& =& -192\left(15+{\left|x\right|}^2\right)\left(1-{\left|x\right|}^2\right){\left(4-c^2\right)}^2,\hfill \\ \hfill {\mathfrak{b}}_3\left(c,x,\delta \right)& =& 3072\left(1-{\left|\delta \right|}^2\right)\left(4-c^2\right)\left(1-{\left|x\right|}^2\right)x.\hfill \end{array}$$

By replacing $\left|\delta \right|$ by y and $\left|x\right|$ by x, if we apply the relation $\left|\rho \right|\le 1$, it follows that

$$\begin{array}{ccc}\hfill \left|a_3{a}_5-a_4^2\right|& \le & \frac{1}{2949120}\left(\left|{\mathfrak{b}}_0\left(c,x\right)\right|+\left|{\mathfrak{b}}_1\left(c,x\right)\right|y+\left|{\mathfrak{b}}_2\left(c,x\right)\right|y^2+\left|{\mathfrak{b}}_3\left(c,x,\delta \right)\right|\right),\hfill \\ & \le & \frac{1}{2949120}\left(\mathfrak{f}\left(c,x,y\right)\right),\hfill \end{array}$$

(37)

where

$$\mathfrak{f}\left(c,x,y\right)={\mathfrak{z}}_0\left(c,x\right)+{\mathfrak{z}}_1\left(c,x\right)y+{\mathfrak{z}}_2\left(c,x\right)y^2+{\mathfrak{z}}_3\left(c,x\right)\left(1-y^2\right)$$

(38)

with

$$\begin{array}{ccc}\hfill {\mathfrak{z}}_0\left(c,x\right)& =& 5c^6+24x^2\left[5c^4+2\left(4c^2+x^2{c}^2\right)\left(4-c^2\right)\right]\left(4-c^2\right),\hfill \\ \hfill {\mathfrak{z}}_1\left(c,x\right)& =& 48c\left[5c^2+4x^2\left(4-c^2\right)\right]\left(1-x^2\right)\left(4-c^2\right),\hfill \\ \hfill {\mathfrak{z}}_2\left(c,x\right)& =& 192\left(15+x^2\right)\left(1-x^2\right){\left(4-c^2\right)}^2,\hfill \\ \hfill {\mathfrak{z}}_3\left(c,x\right)& =& 3072\left(1-x^2\right){\left(4-c^2\right)}^{2}x.\hfill \end{array}$$

Now, we have to maximize $\mathfrak{f}\left(c,x,y\right)$ in the closed cuboid $\Delta :\left[0,2\right]\times \left[0,1\right]\times \left[0,1\right]$. For this purpose, we have to find the maximum of $\mathfrak{f}\left(c,x,y\right)$ in the interior of $\Delta$, in the interior of all its six faces, and on the twelve edges of $\Delta$.

(i). Initially, we will look for the maxima of $\mathfrak{f}\left(c,x,y\right)$ in the interior of $\Delta$.

Let $\left(c,x,y\right)\in \left(0,2\right)\times \left(0,1\right)\times \left(0,1\right)$. Then, on differentiating (38) partially with respect to parameter y, one may obtain

$$\begin{array}{ccc}\hfill \frac{\partial \mathfrak{f}}{\partial y}& =& 48\left(4-c^2\right)\left(1-x^2\right)\left[8\left(x-15\right)\left(4-c^2\right)\left(x-1\right)y\right.\hfill \\ & & \left.+\left(4\left(4-c^2\right)x^2+5c^2\right)c\right].\hfill \end{array}$$

Taking $\frac{\partial \mathfrak{f}}{\partial y}=0$, we obtain

$$y=\frac{\left(4x^2\left(4-c^2\right)+5c^2\right)c}{8\left(15-x\right)\left(4-c^2\right)\left(x-1\right)}:=y_0.$$

For $y_0$ to belong to $(0,1)$, we must have

$$5c^3+4c{x}^2(4-c^2)<8\left(x-1\right)\left(4-c^2\right)\left(15-x\right)$$

(39)

and

$$c^2>4.$$

(40)

Now, in order to find the solutions that meet both inequalities (39) and (40), we see that $c^2>4$, and an easy calculation shows that (39) does not hold for all $x\in \left(0,1\right)$. This implies that we found no optimal point of $\mathfrak{f}$ in $\left(0,2\right)\times \left(0,1\right)\times \left(0,1\right)$.

(ii). Next, we look forward to the maximum of $\mathfrak{f}\left(c,x,y\right)$ in the interior of all six faces of $\Delta$.

If we choose $c=0$, then we obtain

$${\mathfrak{j}}_1\left(x,y\right)=\mathfrak{f}\left(0,x,y\right)=3072\left(\left(x-15\right)\left(x-1\right)y^2+16x\right)\left(1-x^2\right),$$

which shows that there does not exist any point of extrema for $\mathfrak{f}\left(0,x,y\right)$ in $\left(0,1\right)\times \left(0,1\right)$. Taking $c=2$, we obtain

$$\mathfrak{f}\left(2,x,y\right)=320.$$

(41)

Setting $x=0$, we have

$${\mathfrak{j}}_2\left(c,y\right)=5c^6+\left(4-c^2\right)\left(11520y^2+240c^{3}y-2880c^2{y}^2\right)=\mathfrak{f}\left(c,0,y\right).$$

We found no solution for the following system of equations

$$\frac{\partial {\mathfrak{j}}_2}{\partial c}=0,\frac{\partial {\mathfrak{j}}_2}{\partial y}=0,$$

in the interval $\left(0,2\right)\times \left(0,1\right)$. Choosing $x=1$, we obtain

$${\mathfrak{j}}_3\left(c,y\right)=\mathfrak{f}\left(c,1,y\right)=125c^6-1440c^4+3840c^2.$$

(42)

For the critical point, $\frac{\partial {\mathfrak{j}}_3}{\partial c}=0$ gives $c_0\approx 1.3104808$, at which, ${\mathfrak{j}}_3$ attains its maximum. That is,

$${\mathfrak{j}}_3\left(c,y\right)\le 2980.765545.$$

Selecting $y=0$, we have

$$\begin{array}{ccc}\hfill {\mathfrak{j}}_4\left(c,x\right)& =& \mathfrak{f}\left(c,x,0\right)=48c^6{x}^4+5c^6-384c^4{x}^4+72c^6{x}^2-1056c^4{x}^2-3072c^4{x}^3\hfill \\ & & +768c^2{x}^4-49152x+3072c^2{x}^2+24576c^2{x}^3-24576c^{2}x\hfill \\ & & -49152x^3+3072c^{4}x.\hfill \end{array}$$

A computation indicates that the solution does not exist for the following equations:

$$\frac{\partial {\mathfrak{j}}_4}{\partial c}=0,\frac{\partial {\mathfrak{j}}_4}{\partial x}=0$$

in $\left(0,2\right)\times \left(0,1\right)$. Substituting $y=1$, we obtain

$$\begin{array}{ccc}\hfill {\mathfrak{j}}_5\left(c,x\right)& =& \mathfrak{f}\left(c,x,1\right)=5c^6+48c^6{x}^4-192c^5{x}^4+72c^6{x}^2-576c^4{x}^4+432c^5{x}^2\hfill \\ & & +1536c^3{x}^4-3744c^4{x}^2-3072x^4+3072c{x}^2+46080+2304c^2{x}^4\hfill \\ & & +960c^3-240c^5-2496c^3{x}^2-3072c{x}^4+2880c^4-23040c^2\hfill \\ & & -43008x^2+24576c^2{x}^2,\hfill \end{array}$$

which shows that the following system of equations

$$\frac{\partial {\mathfrak{j}}_5}{\partial c}=0,\frac{\partial {\mathfrak{j}}_5}{\partial x}=0.$$

has no optimal solution in $\left(0,2\right)\times \left(0,1\right)$.

(iii). Finally, we now look forward to the maximum of $\mathfrak{f}\left(c,x,y\right)$ at the edges of $\Delta$.

By selecting $y=0$ and $x=0$, we obtain

$${\mathfrak{j}}_6\left(c\right)=\mathfrak{f}\left(c,0,0\right)=5c^6,$$

which gives that

$$\mathfrak{f}\left(c,0,0\right)\le 320.$$

By choosing $y=1$ and $x=0$, we obtain

$${\mathfrak{j}}_7\left(c\right)=\mathfrak{f}\left(c,0,1\right)=5c^6-240c^5+2880c^4+960c^3-23040c^2+46080.$$

Clearly, $\frac{\partial {\mathfrak{j}}_7}{\partial c}<0$ for $\left[0,2\right]$, which indicates that ${\mathfrak{j}}_7\left(c\right)$ is decreasing and achieve its maximum at $c=0$. Thus,

$${\mathfrak{j}}_7\left(c\right)\le 46080.$$

By taking $x=0$ and $c=0$, we obtain

$${\mathfrak{j}}_8\left(y\right)=\mathfrak{f}\left(0,0,y\right)=46080y^2,$$

which shows that

$$\mathfrak{f}\left(0,0,y\right)\le 46080.$$

We note that (42) is free of y. It follows that

$$\begin{array}{ccc}\hfill {\mathfrak{j}}_9\left(c\right)& =& \mathfrak{f}\left(c,1,0\right)=\mathfrak{f}\left(c,1,1\right)\hfill \\ & =& 125c^6-1440c^4+3840c^2.\hfill \end{array}$$

For the critical point, $\frac{\partial {\mathfrak{j}}_9}{\partial c}=0$ gives $c_0=1.310480$, at which, the maximum value is attained for ${\mathfrak{j}}_9\left(c\right)$. Thus,

$${\mathfrak{j}}_9\left(c\right)\le 2980.76554.$$

By setting $x=1$ and $c=0$, we obtain

$$\mathfrak{f}\left(0,1,y\right)=0.$$

By substituting $c=2$, we see that (41) is free from y, x, and c. It follows that

$$\mathfrak{f}\left(2,0,y\right)=\mathfrak{f}\left(2,1,y\right)=\mathfrak{f}\left(2,x,1\right)=\mathfrak{f}\left(2,x,0\right)=320.$$

By choosing $y=0$ and $c=0$, the edge $\mathfrak{f}\left(c,x,y\right)$ yields

$${\mathfrak{j}}_{10}\left(x\right)=\mathfrak{f}\left(0,x,0\right)=-49152x^3+49152x.$$

For the critical point, $\frac{\partial {\mathfrak{j}}_{10}}{\partial x}=0$ gives $x_0=0.577350$, at which, the maximum value is achieved for ${\mathfrak{j}}_{10}\left(x\right)$. Thus,

$$\mathfrak{f}\left(0,x,0\right)\le \frac{32768}{3}\sqrt{3}.$$

By taking $y=1$ and $c=0$, we obtain

$${\mathfrak{j}}_{11}\left(x\right)=\mathfrak{f}\left(0,x,1\right)=-3072(1-x^2)(x^2+15).$$

By simple computation, we see that ${\mathfrak{j}}_{11}\left(x\right)$ obtains its maximum value at 0, so we have

$$\mathfrak{f}\left(0,x,1\right)\le 46080.$$

Hence, from the above situations, we conclude that

$$\mathfrak{f}\left(c,x,y\right)\le 46080\mathrm{on}\Delta :\left[0,2\right]\times \left[0,1\right]\times \left[0,1\right].$$

By using Equation (37), it follows that

$$\left|a_3{a}_5-a_4^2\right|\le \frac{1}{2949120}\left(\mathfrak{f}\left(c,x,y\right)\right)\le \frac{1}{64}.$$

The required inequality is accomplished. □

4. Logarithmic Coefficient Inequalities for the Class ${\mathfrak{BT}}_{\mathfrak{Sg}}$

Theorem 3.

If the function $f\in {\mathfrak{BT}}_{\mathfrak{Sg}}$ assumes the series representation as given in (1), then

$$\left|{\gamma }_1\right|\le \frac{1}{8},\left|{\gamma }_2\right|\le \frac{1}{12},\left|{\gamma }_3\right|\le \frac{1}{16},\left|{\gamma }_4\right|\le \frac{1}{20}.$$

These inequalities are sharp, which can be seen from (4)–(7) and

$$\begin{array}{ccc}\hfill f_0\left(z\right)& =& {\int }_0^z\left(\frac{2}{1+e^{-t}}\right)dt=z+\frac{z^2}{4}-\frac{z^4}{96}+\cdots ,\hfill \\ \hfill f_1\left(z\right)& =& {\int }_0^z\left(\frac{2}{1+e^{-\left(t^2\right)}}\right)dt=z+\frac{z^3}{6}-\frac{z^7}{168}+\cdots ,\hfill \\ \hfill f_2\left(z\right)& =& {\int }_0^z\left(\frac{2}{1+e^{-\left(t^3\right)}}\right)dt=z+\frac{z^4}{8}-\frac{z^{10}}{240}+\cdots ,\hfill \\ \hfill f_3\left(z\right)& =& {\int }_0^z\left(\frac{2}{1+e^{-\left(t^4\right)}}\right)dt=z+\frac{z^5}{10}-\frac{z^{13}}{312}+\cdots .\hfill \end{array}$$

Proof. 

Substituting (24)–(27) into (4)–(7), we obtain

$$\begin{array}{ccc}\hfill {\gamma }_1& =& \left(\frac{1}{16}\right)c_1,\hfill \end{array}$$

(43)

$$\begin{array}{ccc}\hfill {\gamma }_2& =& -\frac{19}{768}{c}_1^2+\frac{1}{24}{c}_2,\hfill \end{array}$$

(44)

$$\begin{array}{ccc}\hfill {\gamma }_3& =& -\frac{7}{192}{c}_1{c}_2+\frac{31}{3072}{c}_1^3+\frac{1}{32}{c}_3,\hfill \end{array}$$

(45)

$$\begin{array}{ccc}\hfill {\gamma }_4& =& \frac{541}{23040}{c}_1^2{c}_2-\frac{37}{1280}{c}_1{c}_3-\frac{5941}{1474560}{c}_1^4+\frac{1}{40}{c}_4-\frac{41}{2880}{c}_2^2.\hfill \end{array}$$

(46)

Applying (14) in (43), we obtain

$$\left|{\gamma }_1\right|\le \frac{1}{8}.$$

Now, from (44), we can write

$${\gamma }_2=\frac{1}{24}\left(c_2-\frac{19}{32}{c}_1^2\right).$$

Applying (13), we obtain

$$\left|{\gamma }_2\right|\le \frac{1}{12}.$$

For (45), we deduce that

$$\left|{\gamma }_3\right|=\frac{1}{32}\left|\left(c_3-2\left(\frac{7}{12}\right)c_1{c}_2+\left(\frac{31}{96}\right)c_1^3\right)\right|.$$

From (15), let

$$B=\frac{7}{12}\mathrm{and}D=\frac{31}{96}.$$

It is clear that $0\le B\le 1$, $B\ge D$ and

$$B\left(2B-1\right)=\frac{7}{72}\le D.$$

Thus, all of the conditions of Lemma (2) are satisfied. Thus, we have

$$\left|{\gamma }_3\right|\le \frac{1}{16}.$$

From (46), we conclude that

$$\begin{array}{ccc}\hfill {\gamma }_4& =& -\frac{1}{40}\left(\frac{5941}{36864}{c}_1^4+\left(\frac{41}{72}\right)c_2^2+2\left(\frac{37}{64}\right)c_1{c}_3-\frac{3}{2}\left(\frac{541}{864}\right)c_1^2{c}_2-c_4\right)\hfill \end{array}$$

(47)

$$\begin{array}{ccc}& =& -\frac{1}{40}\left(\eta c_1^4+a{c}_2^2+2\alpha c_1{c}_3-\frac{3}{2}\beta c_1^2{c}_2-c_4\right),\hfill \end{array}$$

(48)

where

$$\eta =\frac{5941}{36864},a=\frac{41}{72},\alpha =\frac{37}{64},\beta =\frac{541}{864}.$$

We see that $0<a<1$, $0<\alpha <1$, and

$$8a\left(1-a\right)\left({\left(\alpha \beta -2\eta \right)}^2+{\left(\alpha \left(a+\alpha \right)-\beta \right)}^2\right)+\alpha \left(1-\alpha \right){\left(\beta -2a\alpha \right)}^2=0.006067,$$

and

$$4a{\alpha }^2{\left(1-\alpha \right)}^2\left(1-a\right)=0.05833.$$

Thus, the conditions of Lemma 4 are satisfied. Thus, we have

$$\left|{\gamma }_4\right|\le \frac{1}{20}.$$

This completes the proof. □

Theorem 4.

Let the function $f\in {\mathfrak{BT}}_{\mathfrak{Sg}}$ be of the form given in (1). Then,

$$\left|{\gamma }_2-\lambda {\gamma }_1^2\right|\le max\left\{\frac{1}{12},\left|\frac{3+3\left|\lambda \right|}{192}\right|\right\},\mathrm{for}\lambda \in \mathbb{C}.$$

The result is sharp, which can be obtained from (4), (5), and

$$f_1\left(z\right)={\int }_0^z\left(\frac{2}{1+e^{-\left(t^2\right)}}\right)dt=z+\frac{z^3}{6}-\frac{z^7}{168}+\cdots .$$

Proof. 

From (43) and (44), one may write

$$\left|{\gamma }_2-\lambda {\gamma }_1^2\right|=\left|\frac{1}{24}{c}_2-\frac{19}{768}{c}_1^2-\frac{\lambda }{256}{c}_1^2\right|.$$

The application of (13) gives that

$$\left|{\gamma }_2-\lambda {\gamma }_1^2\right|\le \frac{2}{24}max\left\{1,\left|\frac{19+3\lambda }{16}-1\right|\right\}.$$

After simplification, we obtain

$$\left|{\gamma }_2-\lambda {\gamma }_1^2\right|\le max\left\{\frac{1}{12},\left|\frac{3+3\left|\lambda \right|}{192}\right|\right\},$$

which gives the required result. □

Theorem 5.

Let the function $f\in {\mathfrak{BT}}_{\mathfrak{Sg}}$ be of the form given in (1). Then,

$$\left|{\gamma }_1{\gamma }_2-{\gamma }_3\right|\le \frac{1}{16}.$$

The inequality is sharp for (4)–(6) and

$$f_2\left(z\right)={\int }_0^z\left(\frac{2}{1+e^{-\left(t^3\right)}}\right)dt=z+\frac{z^4}{8}-\frac{z^{10}}{240}+\cdots .$$

Proof. 

Using (43–45), we have

$$\left|{\gamma }_1{\gamma }_2-{\gamma }_3\right|=\frac{1}{32}\left|c_3-2\left(\frac{5}{8}\right)c_1{c}_2+\left(\frac{143}{384}\right)c_1^3\right|.$$

From (15), let

$$B=\frac{5}{8}\mathrm{and}D=\frac{143}{384}.$$

Applying Lemma 2, we obtain

$$\left|{\gamma }_1{\gamma }_2-{\gamma }_3\right|\le \frac{1}{16}.$$

□

Theorem 6.

Let the function $f\in {\mathfrak{BT}}_{\mathfrak{Sg}}$ be of the form given in (1). Then,

$$\left|{\gamma }_4-{\gamma }_2^2\right|\le \frac{1}{20}.$$

The result is sharp for (5), (7) and

$$f_3\left(z\right)={\int }_0^z\left(\frac{2}{1+e^{-\left(t^4\right)}}\right)dt=z+\frac{z^5}{10}-\frac{z^{13}}{312}+\cdots .$$

Proof. 

By using (44) and (46), we obtain

$$\begin{array}{ccc}\hfill \left|{\gamma }_4-{\gamma }_2^2\right|& =& \left|-\frac{23}{1440}{c}_2^2+\frac{1177}{46080}{c}_1^2{c}_2-\frac{13687}{2949120}{c}_1^4-\frac{37}{1280}{c}_1{c}_3+\frac{1}{40}{c}_4\right|\hfill \\ & =& \frac{1}{40}\left|\frac{13687}{73728}{c}_1^4+\frac{23}{36}{c}_2^2+2\left(\frac{37}{64}\right)c_1{c}_3-\frac{3}{2}\left(\frac{1177}{1728}\right)c_1^2{c}_2-c_4\right|\hfill \\ & =& \frac{1}{40}\left|\eta c_1^4+a{c}_2^2+2\alpha c_1{c}_3-\frac{3}{2}\beta c_1^2{c}_2-c_4\right|,\hfill \end{array}$$

where

$$\eta =\frac{13687}{73728},a=\frac{23}{36},\alpha =\frac{37}{64},\beta =\frac{1177}{1728}.$$

We see that $0<a<1$, $0<\alpha <1$, and

$$8a\left(1-a\right)\left({\left(\alpha \beta -2\eta \right)}^2+{\left(\alpha \left(a+\alpha \right)-\beta \right)}^2\right)+\alpha \left(1-\alpha \right){\left(\beta -2a\alpha \right)}^2=0.002673,$$

and

$$4a{\alpha }^2{\left(1-\alpha \right)}^2\left(1-a\right)=0.05489.$$

Thus, all of the conditions of Lemma 4 are satisfied. Hence, we have

$$\left|{\gamma }_4-{\gamma }_2^2\right|\le \frac{1}{20}.$$

□

5. Hankel Determinant with Logarithmic Coefficients for the Class ${\mathfrak{BT}}_{\mathfrak{Sg}}$

Theorem 7.

If the function $f\in {\mathfrak{BT}}_{\mathfrak{Sg}}$ assumes the series representation given in (1), then

$$\left|{\mathsf{Ħ}}_{2,1}\left(F_f/2\right)\right|\le \frac{1}{144}.$$

The result is sharp and equality can be achieved from (4)–(6) and

$$f_1\left(z\right)={\int }_0^z\left(\frac{2}{1+e^{-\left(t^2\right)}}\right)dt=z+\frac{z^3}{6}-\frac{z^7}{168}+\cdots .$$

Proof. 

Substituting (43)–(45) into (9), we have

$${\mathsf{Ħ}}_{2,1}\left(F_f/2\right)=-\frac{1}{4608}{c}_1^2{c}_2+\frac{11}{589824}{c}_1^4+\frac{1}{512}{c}_1{c}_3-\frac{1}{576}{c}_2^2.$$

Using (16) and (17) to express $c_2$ and $c_3$ in terms of $c_1$ and setting $c_1=c$, one may have

$$\begin{array}{ccc}\hfill \left|{\mathsf{Ħ}}_{2,1}\left(F_f/2\right)\right|& =& \left|-\frac{1}{2304}{\left(4-c^2\right)}^2{x}^2-\frac{1}{2048}\left(4-c^2\right)x^2{c}^2-\frac{7}{196608}{c}^4\right.\hfill \\ & & \left.+\frac{1}{1024}\left(1-{\left|x\right|}^2\right)\left(4-c^2\right)c\delta \right|,\hfill \end{array}$$

Setting $\left|x\right|=u$, $\left|\delta \right|\le 1$ with $u\le 1$ and applying the triangle inequality, we have

$$\begin{array}{ccc}\hfill \left|{\mathsf{Ħ}}_{2,1}\left(F_f/2\right)\right|& \le & \frac{1}{2304}{\left(4-c^2\right)}^2{u}^2+\frac{1}{2048}\left(4-c^2\right)c^2{u}^2+\frac{7}{196608}{c}^4\hfill \\ & & +\frac{1}{1024}\left(1-u^2\right)\left(4-c^2\right)c:=\mathsf{\Psi }\left(c,u\right).\hfill \end{array}$$

Now, differentiating $\mathsf{\Psi }\left(c,u\right)$ with respect to parameter u, we have

$$\frac{\partial \mathsf{\Psi }\left(c,u\right)}{\partial u}=\left(\frac{1}{9216}{c}^2+\frac{1}{288}-\frac{1}{512}c\right)\left(4-c^2\right)u.$$

We see that $\frac{\partial \mathsf{\Psi }\left(c,u\right)}{\partial u}\ge 0$ on $\left[0,1\right]$, which shows that $\mathsf{\Psi }\left(c,u\right)\le \mathsf{\Psi }\left(c,1\right)$. Thus,

$$\left|{\mathsf{Ħ}}_{2,1}\left(F_f/2\right)\right|\le \frac{1}{2304}{\left(4-c^2\right)}^2+\frac{c^2}{2048}\left(4-c^2\right)+\frac{7c^4}{196608}:=L\left(c\right).$$

By simple computation, it follows that $L\left(c\right)$ obtains its maximum value at $c=0$. Hence,

$$\left|{\mathsf{Ħ}}_{2,1}\left(F_f/2\right)\right|\le \frac{1}{144}.$$

□

Theorem 8.

If the function $f\in {\mathfrak{BT}}_{\mathfrak{Sg}}$ assumes the series representation given in (1), then

$$\left|{\mathsf{Ħ}}_{2,2}\left(F_f/2\right)\right|\le \frac{1}{256}.$$

The result is sharp. Equality can be achieved from (5)–(7) and

$$f_2\left(z\right)={\int }_0^z\left(\frac{2}{1+e^{-\left(t^3\right)}}\right)dt=z+\frac{z^4}{8}-\frac{z^{10}}{240}+\cdots .$$

Proof. 

The determinant ${\mathsf{Ħ}}_{2,2}\left(F_f/2\right)$ is described as follows.

$${\mathsf{Ħ}}_{2,2}\left(F_f/2\right)={\gamma }_2{\gamma }_4-{\gamma }_3^2.$$

By virtue of (44)–(46), along with $c_1=c\in \left[0,2\right]$, it can be written that

$$\begin{array}{ccc}\hfill {\mathsf{Ħ}}_{2,2}\left(F_f/2\right)& =& \frac{1}{T}\left(95616c^3{c}_3+1179648c_2{c}_4-14688c^4{c}_2-671744c_2^3-1105920c_3^2\right.\hfill \\ & & \left.+1216512c{c}_2{c}_3-2441c^6-700416c^2{c}_4+1536c^2{c}_2^2\right),\hfill \end{array}$$

(49)

where $T=1132462080$ and by substituting $m=4-c^2$ in (16)–(18). Now, applying these lemmas, we have

$$\begin{array}{ccc}\hfill 95616c^3{c}_3& =& 23904\left(c^6-c^{4}m{x}^2\right)+47808\left(c^{4}xm+c^{3}m\left(1-{\left|x\right|}^2\right)\delta \right),\hfill \\ \hfill 14688c^4{c}_2& =& 7344c^{4}mx+7344c^6,\hfill \\ \hfill 671744c_2^3& =& 251904\left(c^{4}mx+c^2{m}^2{x}^2\right)+83968\left(m^3{x}^3+c^6\right),\hfill \\ \hfill 1216512c{c}_2{c}_3& =& -152064x^3{m}^2{c}^2-152064c^{4}m{x}^2+304128cx{m}^2\left(1-{\left|x\right|}^2\right)\delta \hfill \\ & & +304128c^2{x}^2{m}^2+304128c^{3}m\left(1-{\left|x\right|}^2\right)\delta +456192c^{4}xm\hfill \\ & & +152064c^6.\hfill \\ \hfill 1179648c_2{c}_4& =& -294912\left(1-{\left|x\right|}^2\right)\overline{x}{\delta }^{2}m{c}^2+73728c^{4}m{x}^3+294912c^{4}xm+294912\hfill \\ & & m{c}^2{x}^2+73728x^4{m}^2{c}^2+221184c^2{x}^2{m}^2-221184c^{4}m{x}^2+73728c^6\hfill \\ & & +294912\left(1-{\left|x\right|}^2\right)c^3\delta m+294912\left(1-{\left|\delta \right|}^2\right)\left(1-{\left|x\right|}^2\right)c^2\rho m\hfill \\ & & -294912\left(1-{\left|x\right|}^2\right)c^3\delta xm-221184x^3{m}^2{c}^2+294912x^3{m}^2\hfill \\ & & -294912x^2{m}^2\left(1-{\left|x\right|}^2\right)c\delta -294912x{m}^2\left(1-{\left|x\right|}^2\right)\overline{x}{\delta }^2+294912\hfill \\ & & m^{2}x\left(1-{\left|\delta \right|}^2\right)\left(1-{\left|x\right|}^2\right)\rho +294912\left(1-{\left|x\right|}^2\right)m^2\delta xc,\hfill \\ \hfill 700416c^2{c}_4& =& 87552c^{4}m{x}^3-350208\left(1-{\left|x\right|}^2\right)\overline{x}{\delta }^{2}m{c}^2+87552c^6+262656c^{4}xm\hfill \\ & & +350208\left(1-{\left|\delta \right|}^2\right)\left(1-{\left|x\right|}^2\right)c^2\rho m+350208m{c}^2{x}^2-262656c^{4}m{x}^2\hfill \\ & & +350208\left(1-{\left|x\right|}^2\right)c^3\delta m-350208\left(1-{\left|x\right|}^2\right)c^3\delta mx,\hfill \\ \hfill 1105920c_3^2& =& -276480c{x}^2{m}^2\left(1-{\left|x\right|}^2\right)\delta +69120x^4{m}^2{c}^2-138240c^{4}m{x}^2\hfill \\ & & -276480x^3{m}^2{c}^2+276480c^2{x}^2{m}^2+552960cx{m}^2\left(1-{\left|x\right|}^2\right)\delta \hfill \\ & & +276480m^2{\left(1-{\left|x\right|}^2\right)}^2{\delta }^2+276480c^{4}xm+276480c^{3}m\left(1-{\left|x\right|}^2\right)\delta \hfill \\ & & +69120c^6,\hfill \\ \hfill 1536c^2{c}_2^2& =& 768c^{4}mx+384\left(c^6+c^2{m}^2{x}^2\right).\hfill \end{array}$$

Inserting the above expression into (49) , we achieve

$$\begin{array}{ccc}\hfill {\mathsf{Ħ}}_{2,2}\left(F_f/2\right)& =& \frac{1}{T}\left\{55296c^{3}m\left(1-{\left|x\right|}^2\right)x\delta +55296c^{2}m\left(1-{\left|x\right|}^2\right)\overline{x}{\delta }^2\right.\hfill \\ & & -55296c^{2}m\left(1-{\left|x\right|}^2\right)\left(1-{\left|\delta \right|}^2\right)\rho +46080cx{m}^2\left(1-{\left|x\right|}^2\right)\delta \hfill \\ & & -18432x^2{m}^2\left(1-{\left|x\right|}^2\right)c\delta -294912x{m}^2\left(1-{\left|x\right|}^2\right)x{\delta }^2+294912x\hfill \\ & & m^2\left(1-{\left|x\right|}^2\right)\left(1-{\left|\delta \right|}^2\right)\rho +1296c^{4}xm+3744c^{4}m{x}^2-2688c^2{x}^2{m}^2\hfill \\ & & -55296m{c}^2{x}^2-13824c^{4}m{x}^3-96768x^3{m}^2{c}^2+4608x^4{m}^2{c}^2\hfill \\ & & -276480m^2{\left(1-{\left|x\right|}^2\right)}^2{\delta }^2+20160c^{3}m\left(1-{\left|x\right|}^2\right)\delta -83968x^3{m}^3\hfill \\ & & \left.+294912x^3{m}^2-345c^6\right\}.\hfill \end{array}$$

Since $m=4-c^2$, it follows that

$${\mathsf{Ħ}}_{2,2}\left(F_f/2\right)=\frac{1}{T}\left({\mathfrak{h}}_1\left(c,x\right)+{\mathfrak{h}}_2\left(c,x\right)\delta +{\mathfrak{h}}_3\left(c,x\right){\delta }^2+\zeta \left(c,x,\delta \right)\rho \right),$$

where $\delta ,x,\rho \in \overline{O_d}$, and

$$\begin{array}{ccc}\hfill {\mathfrak{h}}_1\left(c,x\right)& =& -345c^6+\left(4-c^2\right)\left[\left(4-c^2\right)\left(4608c^2{x}^4-40960x^3-12800c^2{x}^3\right.\right.\hfill \\ & & \left.\left.-2688c^2{x}^2\right)-13824c^4{x}^3+1296c^{4}x+3744c^4{x}^2-55296c^2{x}^2\right],\hfill \\ \hfill {\mathfrak{h}}_2\left(c,x\right)& =& \left(4-c^2\right)\left(1-{\left|x\right|}^2\right)\left[\left(4-c^2\right)\left(-18432c{x}^2+46080cx\right)+55296c^{3}x+20160c^3\right],\hfill \\ \hfill {\mathfrak{h}}_3\left(c,x\right)& =& \left(4-c^2\right)\left(1-{\left|x\right|}^2\right)\left[\left(4-c^2\right)\left(-276480-18432{\left|x\right|}^2\right)+55296c^2\overline{x}\right],\hfill \\ \hfill \zeta \left(c,x,\delta \right)& =& \left(4-c^2\right)\left(1-{\left|x\right|}^2\right)\left(1-{\left|\delta \right|}^2\right)\left[-55296c^2+294912x\left(4-c^2\right)\right].\hfill \end{array}$$

By replacing $\left|\delta \right|$ with y and $\left|x\right|$ with x, if we apply the inequality $\left|\rho \right|\le 1$, it follows that

$$\begin{array}{ccc}\hfill \left|{\mathsf{Ħ}}_{2,2}\left(F_f/2\right)\right|& \le & \frac{1}{T}\left(\left|{\mathfrak{h}}_1\left(c,x\right)\right|+y\left|{\mathfrak{h}}_2\left(c,x\right)\right|+y^2\left|{\mathfrak{h}}_3\left(c,x\right)\right|+\left|\zeta \left(c,x,\delta \right)\right|\right).\hfill \\ & \le & \frac{1}{T}\left(\mathcal{W}\left(c,x,y\right)\right),\hfill \end{array}$$

(50)

where

$$\mathcal{W}\left(c,x,y\right)={\mathfrak{r}}_0\left(c,x\right)+{\mathfrak{r}}_1\left(c,x\right)y+{\mathfrak{r}}_2\left(c,x\right)y^2+{\mathfrak{r}}_3\left(c,x\right)\left(1-y^2\right),$$

with

$$\begin{array}{ccc}\hfill {\mathfrak{r}}_0\left(c,x\right)& =& 345c^6+\left[\left(4-c^2\right)\left(40960x^3+4608c^2{x}^4+12800c^2{x}^3+2688c^2{x}^2\right)\right.\hfill \\ & & \left.+3744c^4{x}^2+13824c^4{x}^3+55296c^2{x}^2+1296c^{4}x\right]\left(4-c^2\right),\hfill \\ \hfill {\mathfrak{r}}_1\left(c,x\right)& =& \left(4-c^2\right)\left(1-x^2\right)\left[\left(4-c^2\right)\left(18432c{x}^2+46080cx\right)+55296c^{3}x+20160c^3\right],\hfill \\ \hfill {\mathfrak{r}}_2\left(c,x\right)& =& \left(4-c^2\right)\left(1-x^2\right)\left[\left(4-c^2\right)\left(276480+18432x^2\right)+55296c^{2}x\right],\hfill \\ \hfill {\mathfrak{r}}_3\left(c,x\right)& =& \left(4-c^2\right)\left(1-x^2\right)\left[55296c^2+294912x\left(4-c^2\right)\right].\hfill \end{array}$$

Now, we have to maximize $\mathcal{W}\left(c,x,y\right)$ in the closed cuboid $\Delta :\left[0,2\right]\times \left[0,1\right]\times \left[0,1\right]$. For this purpose, we have to find the maxima of $\mathcal{W}\left(c,x,y\right)$ in the interior of $\Delta$, in the interior of its six faces, and on the twelve edges of $\Delta$.

(i). Initially, we find the maximum of $\mathcal{W}\left(c,x,y\right)$ in the interior of $\Delta$.

Let $\left(c,x,y\right)\in \left(0,2\right)\times \left(0,1\right)\times \left(0,1\right)$. Then, when differentiating $\mathcal{W}\left(c,x,y\right)$ partially with respect to parameter y, it implies that

$$\begin{array}{ccc}\hfill \frac{\partial \mathcal{W}}{\partial y}& =& 576(1-x^2)\left(4-c^2\right)\left[\left(192c^2+64\left(4-c^2\right)\left(x-15\right)\right)(x-1)y\right.\hfill \\ & & \left.+\left(c^2\left(35+96x\right)+x\left(4-c^2\right)\left(80+32x\right)\right)c\right].\hfill \end{array}$$

The equation $\frac{\partial \mathcal{W}}{\partial y}=0$ gives

$$y=\frac{\left(c^2\left(35+96x\right)+x\left(4-c^2\right)\left(80+32x\right)\right)c}{\left(-192c^2+64\left(4-c^2\right)\left(15-x\right)\right)(x-1)}:=y_3.$$

For $y_3$ to belong to $\left(0,1\right)$, it must follow that

$$c^3\left(35+96x\right)+cx\left(80+32x\right)\left(4-c^2\right)+64(1-x)\left(15-x\right)\left(4-c^2\right)<192c^2(1-x).$$

(51)

and

$$c^2>\frac{3840-256x}{1152-64x}.$$

(52)

Now, in order to find the solutions that meet both inequalities (51) and (52), we see that $v\left(x\right)=\frac{3840-256x}{1152-64x}$ gives that $v^{\prime }\left(x\right)<0$ for $\left(0,1\right)$, which shows that $v\left(x\right)$ is decreasing over $\left(0,1\right)$. Thus, $c^2>\frac{56}{17}$ and an easy calculation shows that (51) does not hold for all $x\in \left(0,1\right)$. Therefore, there are no critical points of $\mathcal{W}$ in the interval $\left(0,2\right)\times \left(0,1\right)\times \left(0,1\right)$.

(ii). Next, we look forward to find the maximum of $\mathcal{W}\left(c,x,y\right)$ in the interior of all six faces of $\Delta$.

Choosing $c=0$, we obtain

$$\begin{array}{ccc}\hfill {\mathfrak{t}}_1(x,y)& =& \mathcal{W}(0,x,y)=-4063232x^3+294912(x-1)(x-15)(1-x^2)y^2\hfill \\ & & +4718592x,\hfill \end{array}$$

which shows that there does not exist any $max\mathcal{W}(0,x,y)$ in $\left(0,1\right)\times \left(0,1\right)$. Taking $c=2$, we have

$$\mathcal{W}(2,x,y)=22080.$$

(53)

Selecting $x=0$, we find that

$$\begin{array}{ccc}\hfill {\mathfrak{t}}_2(c,y)& =& \mathcal{W}\left(c,0,y\right)=345c^6+(4-c^2)\left(20160c^{3}y-331776c^2{y}^2+55296c^2\right.\hfill \\ & & \left.+1105920y^2\right).\hfill \end{array}$$

For the critical points, we see that $\frac{\partial {\mathfrak{t}}_2}{\partial y}=0$ gives

$$y=\frac{35c^3}{384(3c^2-10)}:=y_0.$$

(54)

For the concerned range of y, $y_0$ would belong to $\left(0,1\right)$ only if $c>1.880236$. Furthermore,

$$\begin{array}{ccc}\hfill \frac{\partial {\mathfrak{t}}_2}{\partial c}& =& \left(4-c^2\right)\left(60480c^{2}y-663552c{y}^2+110592c\right)+2070c^5-40320c^{4}y\hfill \\ & & +663552c^3{y}^2-110592c^3-2211840c{y}^2.\hfill \end{array}$$

Inserting (54) into the above expression and setting $\frac{\partial {\mathfrak{t}}_2}{\partial c}=0$, we obtain

$$\frac{\partial {\mathfrak{t}}_2}{\partial c}=4185c^9-3994512c^7+34477704c^5-97320960c^3+88473600c=0.$$

Now, solving for $c\in \left(0,2\right)$, we obtain $c\approx 1.4072104$. Thus, no optimal solution is achieved for $\mathcal{W}\left(c,0,y\right)$ in $\left(0,2\right)\times \left(0,1\right)$. If we choose $x=1$, we obtain

$${\mathfrak{t}}_3(c,y)=\mathcal{W}(c,1,y)=1577c^6-99648c^4+215040c^2+655360.$$

(55)

For the critical point, $\frac{\partial {\mathfrak{t}}_3}{\partial c}=0$ gives $c_0\approx 1.052686$, at which, ${\mathfrak{t}}_3$ attains its maximum value; that is,

$$\mathcal{W}(c,1,y)\le 773435.177.$$

Setting $y=0$, we have

$$\begin{array}{ccc}\hfill {\mathfrak{t}}_4(c,x)& =& \mathcal{W}(c,x,0)=-1024c^6{x}^3+4608c^6{x}^4-36864c^4{x}^4-1056c^6{x}^2+2236416c^2{x}^3\hfill \\ & & -301056c^4{x}^3-6528c^4{x}^2+300096c^{4}x+73728c^2{x}^4-1296c^{6}x\hfill \\ & & +43008c^2{x}^2-55296c^4+345c^6-2359296c^{2}x+221184c^2\hfill \\ & & +4718592x-4063232x^3.\hfill \end{array}$$

A computation shows that a solution for a system of the following equations

$$\frac{\partial {\mathfrak{t}}_4}{\partial c}=0\mathsf{\&}\frac{\partial {\mathfrak{t}}_4}{\partial x}=0$$

in $\left(0,2\right)\times \left(0,1\right)$ does not exist.

Substituting $y=1$, we find that

$$\begin{array}{ccc}\hfill {\mathfrak{t}}_5(c,x)& =& \mathcal{W}(c,x,1)=-1024c^6{x}^3+4608c^6{x}^4-1056c^6{x}^2-18432c^5{x}^4+9216c^5{x}^3+345c^6\hfill \\ & & -1296c^{6}x-55296c^4{x}^4+49152c^4{x}^3+38592c^5{x}^2+655360x^3\hfill \\ & & -9216c^{5}x+147456c^3{x}^4+147456c^3{x}^3-319872c^4{x}^2+294912c{x}^2\hfill \\ & & -20160c^5+221184c^2{x}^4-228096c^3{x}^2-50112c^{4}x-2211840c^2\hfill \\ & & -344064c^2{x}^3-294912c{x}^4-344064c^2{x}^3+2328576c^2{x}^2\hfill \\ & & -294912x^4-737280c{x}^3+221184c^{2}x+80640c^3+737280cx\hfill \\ & & +4423680+276480c^4-4128768x^2-147456c^{3}x.\hfill \end{array}$$

In this situation, we came to the same conclusion as for ${\mathfrak{t}}_4$; that is, the system

$$\frac{\partial {\mathfrak{t}}_5}{\partial c}=0,\frac{\partial {\mathfrak{t}}_5}{\partial x}=0$$

has no solution in $\left(0,2\right)\times \left(0,1\right)$.

(iii). Finally, we now intend to find the maximum value of $\mathcal{W}\left(c,x,y\right)$ at the twelve edges of $\Delta$. For this, we proceed as follows.

By choosing $y=0$ and $x=0$, it yields that

$${\mathfrak{t}}_6\left(c\right)=\mathcal{W}(c,0,0)=345c^6-55296c^4+221184c^2.$$

For the critical point, $\frac{\partial {\mathfrak{t}}_6}{\partial c}=0$ gives $c_0\approx 1.4279024$, at which, the maximum value is achieved for ${\mathfrak{t}}_6\left(c\right)$. That is,

$$\mathcal{W}(c,0,0)\le \frac{22402453}{100}.$$

By selecting $y=1$ and $x=0$, we obtain

$$\begin{array}{ccc}\hfill {\mathfrak{t}}_7\left(c\right)& =& \mathcal{W}(c,0,1)=-20160c^5+345c^6+80640c^3+276480c^4+276480c^4\hfill \\ & & +4423680+276480c^4.\hfill \end{array}$$

By simple computation, we see that ${\mathfrak{t}}_7\left(c\right)$ obtains its maximum value at 0, so we have

$$\mathcal{W}(c,0,1)\le 4423680.$$

By setting $x=0$ and $c=0$, we have

$${\mathfrak{t}}_8\left(y\right)=\mathcal{W}(0,0,y)=4423680y^2.$$

It follows that $\frac{\partial {\mathfrak{t}}_8}{\partial y}>0$ for $\left[0,1\right]$ shows that ${\mathfrak{t}}_8\left(y\right)$ is decreasing and its maxima is achieved at 1. Hence, we have

$$\mathcal{W}(0,0,y)\le 4423680.$$

We note that (55) is free from y. It follows that

$$\begin{array}{ccc}\hfill {\mathfrak{t}}_9\left(c\right)& =& \mathcal{W}(c,1,1)=\mathcal{W}(c,1,0)\hfill \\ & =& 1577c^6-99648c^4+215040c^2+655360.\hfill \end{array}$$

For the critical point, $\frac{\partial {\mathfrak{t}}_9}{\partial c}=0$ gives $c_0\approx 1.052686$, at which, the maximum value is achieved for ${\mathfrak{t}}_9\left(c\right)$. We conclude that

$$\mathcal{W}(c,1,0)\le 773435.177.$$

By substituting $x=1$ and $c=0$, we obtain

$$\mathcal{W}(0,1,y)=655360.$$

By selecting $c=2$, we see that (53) is free from y, x, and c. It follows that

$$\mathcal{W}(2,1,y)=\mathcal{W}(2,0,y)=\mathcal{W}(2,x,1)=\mathcal{W}(2,x,0)=22080.$$

By taking $y=1$ and $c=0$, we obtain

$${\mathfrak{t}}_{10}\left(x\right)=\mathcal{W}\left(0,x,1\right)=-294912x^4+655360x^3-4128768x^2+4423680.$$

By simple computation, we see that ${\mathfrak{t}}_{10}\left(x\right)$ obtains its maximum value at 0, so we have

$${\mathfrak{t}}_{10}\left(x\right)\le 4423680.$$

By choosing $y=0$ and $c=0$, we have

$${\mathfrak{t}}_{11}\left(x\right)=\mathcal{W}\left(0,x,0\right)=-4063232x^3+4718592x.$$

For the critical point, $\frac{\partial {\mathfrak{t}}_{11}}{\partial x}=0$ gives $x_0\approx 0.6221710$, at which, the maximum value is achieved for ${\mathfrak{t}}_{11}\left(x\right)$. Therefore, we have

$$\mathcal{W}\left(0,x,0\right)\le \frac{6291456}{31}\sqrt{3}\sqrt{31}.$$

Hence, from the above situations, we conclude that

$$\mathcal{W}\left(c,x,y\right)\le 4423680\mathrm{on}\Delta :\left[0,2\right]\times \left[0,1\right]\times \left[0,1\right].$$

By using Equation (50), it follows that

$$\left|{\mathsf{Ħ}}_{2,2}\left(F_f/2\right)\right|\le \frac{1}{T}\left(\mathcal{W}\left(c,x,y\right)\right)\le \frac{1}{256},$$

which completes the proof. □

6. Conclusions

We have obtained the sharp bounds of Hankel determinants of order three for the class of functions with bounding turning that are associated with the sigmoid function. All of the bounds that we found here were sharp. Moreover, we investigated the sharp bounds of logarithmic coefficients linked with the functions of bounded turnings. This also includes the third-order Hankel determinant for these logarithmic coefficients. This work will help in finding the fourth-order Hankel determinants for the same types of analytic functions that have been considered in this study.